The S3 group is the group of permutation on 3 elements. It has 6 elements. It is the group of all the automorphism on the set {1, 2, 3}.
Here is the full list of the elements of S3:
## e: ()
## a: (1,2)
## b: (1,3)
## c: (2,3)
## p: (1,2,3)
## p^2: (1,3,2)
Table of S3:
| S3 | e | p | p2 | a | b | c |
|---|---|---|---|---|---|---|
| e | e | p | p2 | a | b | c |
| p | p | p2 | e | c | a | b |
| p2 | p2 | e | p | b | c | a |
| a | a | b | c | e | p | p2 |
| b | b | c | a | p2 | e | p |
| c | c | a | b | p | p2 | e |
Order of the elements:
| S3 | Order | Sign | Inversions |
|---|---|---|---|
| e | 1 | + | 0 |
| p | 3 | + | 0 |
| p2 | 3 | + | 0 |
| a | 2 | - | 1 |
| b | 2 | - | 1 |
| c | 2 | - | 1 |
Unique normal chain decomposition of S3:
\[ C(2) \lhd C(3) \lhd S_3\]
And this decomposition is unique, the factors cannot be swapped.
grViz('
graph cycle {
graph[layout = neato]
node [shape = circle, style = filled, color = grey, label=""]
e [color = red]; a; b; c; p; p2
e -- a
e -- b
e -- c
e -- p -- p2 -- e
}
')
is a normal subgroup
The group <p> = {e, p, p2} is normal in S3:
all(p^a == p2, p^b == p2, p^c == p2)
## [1] TRUE
The classes of <p> in S3 are:
{e, p, p2}, {a, b, c}
The action of a on <p>: (p p2)
More detailed action of the group on <p>:
| S3 | {e, p, p2} | {a, b, c} |
|---|---|---|
| e | {e, p, p2} | {a, b, c} |
| a | {a, b, c} | {e, p, p2} |
| b | {b, c, a} | {p2, e, p} |
| c | {c, a, b} | {p, p2, e} |
| p | {p, p2, e} | {c, a, b} |
| p2 | {p2, e, p} | {b, c, a} |
<p> is isomorphic to Z/3Z and it has no non-trivial subgroups.
If G is a subgroup of S3 containing <p> then G*<p> is a subgroup of S3/<p>. S3/<p> is a group of order 2 so it is isomorphic to Z/2Z and it has no non trivial subgroup. The only subgroups containing p are {e} and S3.
If G is a sub-group not including <p>. Then inter(G, <p>) is {e}. because inter(G, <p>) is a subgroup of <p> and it cannot be <p> because G does no include <p>.
So G is included in {e, a, b, c}. G cannot contain more than 2 elements because any of products ab, bc, … will yield an element of <p>. So G is either {e, a}, {e, b}, {e, c}.
Conclusion: Here is the full list of the subgroups of S3: {e}, {e, a}, {e, b}, {e, c}, {e, p, p2}, {e, a, b, c, p, p2}
Lattice of all the subgroups
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Normal subgroups
This is a faithful action.
We can compute the orbits:
| S3 | Orbits |
|---|---|
| e | {1} {2} {3} |
| a | {1, 2} {3} |
| b | {1, 3} {2} |
| c | {1} {2, 3} |
| p | {1, 2, 3} |
| p2 | {1, 2, 3} |
We can also compute the stabilizers:
| {1, 2, 3} | Stabilizer |
|---|---|
| 1 | <c> |
| 2 | <b> |
| 3 | <a> |
We consider the action of S3 on this set: {{1, 2}, {1, 3}, {2, 3}}
## [1] TRUE
all(
a_f(c(1, 2)) == c(2, 1),
a_f(c(1,3)) == c(2, 3),
a_f(c(2, 3)) == c(1, 3))
## [1] TRUE
Since a and p generate S3 we deduce that the action is faithfull.
Orbits
| S3 | Orbits |
|---|---|
| e | {1_2} {1_3} {2_3} |
| a | {1_2} {1_3, 2_3} |
| b | {1_2, 2_3} {1_3} |
| c | {1_2, 1_3} {2, 3} |
| p | {1_2, 2_3, 1_3} |
| p2 | {1_2, 2_3, 1_3} |
Stabilizers
The elements of order 2 are: a, b, c
c(a^p == c, a^(p^2) == b)
## [1] TRUE TRUE
So the action is transitive. And |O(a)|.|Stab(a)| = 6 => 3.|Stab| = 6 So the stabilizer size is 2.
And:
| S3 | Stab |
|---|---|
| a | e, a |
| b | e, b |
| c | e, c |
The elements of order 3 are: p, p^2
c(p^a == p^2)
## [1] TRUE
So the action is transitive and: |O(p)|.|Stab(p)| = 6 so |Stab(p)| = 2
| S3 | Stab |
|---|---|
| p | e, p, p2 |
| p2 | e, p, p2 |
The action of S3 on S3/<p> is not faithfull. S3/<p> = {{e, p, p2}, {a, b, c}} Any element of <p> is sent to the identity. Any other element is sent to ({e, p, p2} {a, b, c})
There is also an action of S3 on {a, b, c}.
And an action of S3 on <p>
Let \(\phi\) be a transitive, faithfull set action \(S3 \rightarrow S(\mathcal{A})\). Where \(\mathcal{A}\) is any set. We note: \(g \bullet x = \phi(g)(x)\) for \(g \in S3, a \in \mathcal{A}\)
First we would like to show that \(\mathcal{A}\) has 6 elements.
Let \(a_0 \in \mathcal{A}\), we let \(f_{a_0}: S3 \rightarrow \mathcal{A}\) defined by \(f_{a_0}(g) = g \bullet a_0\)
Since the action is transitive \(f_{a_0}\) is surjective. Let’s \(g_1, g_2 \in \mathrm{S3}\). \[f_{a_0}(g_1) = f_{a_0}(g_2) \iff g_1 \bullet a_0 = g_2 \bullet a_0 \iff g_1 g_2^{-1} \bullet a_0 = a_0\]
In all this file G is a finite group and S is a set.
An action is a group morphism \(\phi_1: G \rightarrow Aut(S)\) where S is some set. Let \(\phi_2: G \rightarrow Aut(S2)\) be another group action. 2 action are isomorphic iff there exist a set isomorphism \(f: S \rightarrow S2\) such that \[\forall g \in G, f(\phi_1(g)(x)) = \phi_2(g)(f(x))\]
The next property shows that if we have verified the property on f for \(g_1, g_2 \in G\), then the property holds for \(g_1g_2\). \[f(\phi_1(g_1g_2)(x)) = f(\phi_1(g_1)(\phi_1(g_2)(x))) = \phi_2(g_1)(f(\phi_1(g_2)x)) = \phi_2(g_1)(\phi_2(g_2)(fx)) = \phi_2(g_1g_2)(fx)\]
That means that we only need to verify the property on a set of generators for G.
Let’s show that the action on {1, 2, 3} is isomorphic to the action on {{1, 2}, {1, 3}, {2, 3}} by taking f: 1 -> {2, 3} f: 2 -> {1, 3} f: 3 -> {1, 2}
f(p.1) = f(2) = {1, 3} p.f(1) = p.{2, 3} = {1, 3}
f(p.2) = f(3) = {1, 2} p.f(2) = p.{1, 3} = {1, 2}
f(a.1) = f(2) = {1, 3} a.f(1) = a.{2, 3} = {1, 3}
f(a.3) = f(3) = a.f(3) => f(3) = {1, 2}
f(b.2) = f(2) = b.f(2) => f(2) = {1, 3}
f(c.1) = f(1) = c.f(1) => f(1) = {2, 3}
So f is an isomorphism of actions.
Conjugation by an element creates an automorphism. The conjugation by p has the following effect:
## a^p: (2,3)
## b^p: (1,2)
## c^p: (1,3)
## p^p: (1,2,3)
## p2^p: (1,3,2)
So conj(p) = (a c b)
Here is the conjugation by p2:
## a^p2: (1,3)
## b^p2: (2,3)
## c^p2: (1,2)
## p^p2: (1,2,3)
## p2^p2: (1,3,2)
So conj(p2) = (a b c)
Here is the conjugation by a:
## a^a: (1,2)
## b^a: (2,3)
## c^a: (1,3)
## p^a: (1,3,2)
## p2^a: (1,2,3)
So conj(a) = (b c) (p p2)
Here is the conjugation by b:
## a^b: (2,3)
## b^b: (1,3)
## c^b: (1,2)
## p^b: (1,3,2)
## p2^b: (1,2,3)
So conj(b) = (a c) (p p2)
Here is the conjugation by c:
## a^c: (1,3)
## b^c: (1,2)
## c^c: (2,3)
## p^c: (1,3,2)
## p2^c: (1,2,3)
So conj(c) = (a b)(p p2)
So together with the identity there are 6 inner automorphisms.
We will show that Inner(S3) = S3 conj(a)^2 = e conj(p)^3 = e conj(p)^conj(a) = conj a-1 o conj p o conj a = conj a-1pa = conj p2 = conj(p)^2
Since the generators of Inner(S3) have the same relations as those of S3 we know the 2 groups are isomorphic.
We can show that there are no outer isomorphism. Indeed an automorphism maps elements to elements of the same order. There are thus 2 possibilities for mapping p and 3 possibilities for mapping a. Since a and p generate S3 this 2 images will define uniquely a mapping. There are at most 3*2 automorphisms and according to the previous paragraphs we already have 6 inner automorphisms. So all the automorphism are inner.
Since an automorphism is uniquely defined by its image of the generators a and p we might be able to write them as a matrix. Indeed the reason we are able to write linear operators as matrix is because they are defined by their action on a basis which is a generator set of the whole vector space.
First lets try to write elements of S3 as vectors: we note a = (1, 0) and p = (0, 1) (0, x) * (1, m) = p^x * a * p^m = a * a p^x * a p^m = (1, 2*x+m)
(g1, n1) * (g2, n2) = g1n1g2n2 = g1g2inv(g2)n1g2n2 = g1g2n1^g2n2 = (g1g2, n1^g2*n2)
inv(g1) * (g2, n2) * g1 = inv(g1) * (g2*g1, n2^g1) = (g2^g1, n2^g1) conj(g1) = g1 e e g1